Extremal Functions for a Mean Field Equation in Two …chang/ccl-20017.pdfExtremal Functions for a...

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Extremal Functions for a Mean Field Equation in Two Dimension Sun-Yung A. Chang, Chiun-Chuan Chen and Chang-Shou Lin Abstract Let Ω be a bounded piecewise C 2 simply-connected domain. In this article, we give necessary and sufficient conditions for the existence of maximizer of J 8π (φ) = log Z - Ω e φ dx ! - 1 16π Z Ω |5 φ| 2 dx for φ H 1 0 (Ω). We prove among other things that the maximizer exists provided that the regular part γ (x) of the Green function for the Dirichlet problem has more than one maximum points in Ω. 1 Introduction Let Ω be a bounded domain in R 2 and ρ be a constant of real numbers. In this paper, we consider the following nonlinear elliptic equation ( Δu + ρ e u R Ω e u dx =0 in Ω, u =0 on Ω, (1.1) where Δ = 2 i=1 2 ∂x 2 i stands for the Laplacian operator of R 2 . Throughout the paper, the domain Ω is assumed to be piecewisely C 2 . A domain Ω is called picewisely C 2 if its boundary is C 2 except at a finite vertex points {Q 1 ,...,Q N } such that two conditions holds at each vertex Q j . (i) At each Q j , the inner angle θ j of Ω at Q j satisfies 0 j 6= π< 2π, 1

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Page 1: Extremal Functions for a Mean Field Equation in Two …chang/ccl-20017.pdfExtremal Functions for a Mean Field Equation in Two Dimension Sun-Yung A. Chang, Chiun-Chuan Chen and Chang-Shou

Extremal Functions for a Mean Field Equationin Two Dimension

Sun-Yung A. Chang, Chiun-Chuan Chen and Chang-Shou Lin

Abstract

Let Ω be a bounded piecewise C2 simply-connected domain. In thisarticle, we give necessary and sufficient conditions for the existence ofmaximizer of

J8π(φ) = log

(

−Ωeφdx

)

− 1

16π

Ω| 5 φ|2dx

for φ ∈ H10 (Ω). We prove among other things that the maximizer

exists provided that the regular part γ(x) of the Green function forthe Dirichlet problem has more than one maximum points in Ω.

1 Introduction

Let Ω be a bounded domain in R2 and ρ be a constant of real numbers.In this paper, we consider the following nonlinear elliptic equation

∆u+ ρ eu∫

Ωeudx

= 0 in Ω,

u = 0 on ∂Ω,(1.1)

where ∆ =∑2i=1

∂2

∂x2i

stands for the Laplacian operator of R2. Throughout

the paper, the domain Ω is assumed to be piecewisely C2. A domain Ω iscalled picewisely C2 if its boundary is C2 except at a finite vertex pointsQ1, . . . , QN such that two conditions holds at each vertex Qj.

(i) At each Qj, the inner angle θj of ∂Ω at Qj satisfies 0 < θj 6= π < 2π,

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(ii) At each Qj, there is a 1-1 conformal map form Bδ0(Qj) ∩ Ω to thecomplex plane C such that the portion ∂Ω ∩ Bδ0(Qj) of the boundaryis mapped onto a C2 curve.

Clearly, non-smooth domains such as n-polygons satisfies conditions (i) and(ii) above. Hence, our theory developed in this article can be applied ton-polygons. See examples in section 6.

Equation (1.1) often appears in many different disciplines of mathematics.Recently, it has been derived in the context of statistical mechanics from themean field limit of the Gibbs measure associated to vorticity fields of Eulerflows, as studied by Caglioti, Lions, Marchioro and Pulvirenti [5, 6], and alsoby Kiessling [18], and Chanillo and Kiessling [8]. A related problem alsoappears in the self-dual condensate solutions of Chern-Simons Higgs modelof superconductivity. See Ding, Jost, Li and Wang [15, 17], Nolasco andTarantello [25, 26] and references therein. In many examples of application,one is interested in the case when ρ = 8π.

Associated with (1.1) is the nonlinear functional Jρ:

Jρ(φ) = log

(

−Ωeφdx

)

− 1

Ω| 5 φ|2dx

for φ ∈ H10 (Ω). It is well-known that Jρ(φ) is bounded above for φ ∈ H1

0(Ω)if and only if ρ ≤ 8π. For ρ < 8π, the supreman of Jρ is always attainedby virtue of the Moser-Trudinger inequality [23]. However, for ρ = 8π, theexistence of extremal functions, which attains the supreman of J8π, is a moredifficult problem and depends on the geometry of Ω in a subtle way. It wasnoted in [5, 6, 7] that when Ω is a ball, the supreman of J8π is never attainedby a function φ in H1

0 (Ω). On the other hand, the extremal functions existsfor a long and thin rectangle Ω. As far as the authors know, there are veryfew known results concerning either the existence of solutions of (1.1) withρ = 8π or the existence of the extremal functions of J8π. Even now, thefollowing basic question is not yet answered. For the simplicity, a domain Ωis called type C if the supreman of J8π can be attained.

[Q]. The domains of type C is open in the C1-topology.

In this paper, we will give an affirmative answer to [Q] at least for simply-connected domains.

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For any bounded smooth domain Ω, we set

I8π(Ω) = supφ∈H1

0 (Ω)

J8π(φ).(1.2)

In the case when the domain is not of type C, we can compute J8π(Ω) ex-plicitly, as done in [5, 6]. We denote G(x, y) to be the Green function andG(x, y) to be the regular part of G:

G(x, y) = G(x, y) +1

2πlog |x− y|.(1.3)

Setγ(x) = G(x, x).(1.4)

When there are no confusions, γ(x) is sometimes called the regular part ofthe Green function. Now suppose Ω is not of type C. Then in [5], I8π(Ω) wascomputed and its value is equal to

I8π(Ω) = 1 + 4π supΩ

γ(x) + log|B1||Ω| .(1.5)

where B1 is the unit ball and |Ω| stands for the area of Ω. See Theorem 7.1of [5] for a proof of (1.5). Surprisingly, the converse holds.

Theorem 1.1. Let Ω be a bounded piecewise C2 simply-connected domain.Then Ω is of type C if and only if

I8π(Ω) > 1 + 4π supΩ

γ(x) + log|B1||Ω| .(1.6)

Readily, Theorem 1.1 give a positive answer to [Q] for simply-connecteddomains.

Corollary 1.2. The simply-connected domains of type C is open in the C1

topology.

We note that the ”if” part of Theorem 1.1 has already been proved in [5].The crucial step of our proof of the ”only if” part is to study the behavior ofa sequence of blowup solutions uk of (1.1) with ρ = ρk, where ρk satisfies

limk→+∞

ρk = 8π.(1.7)

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Under the condition (1.7), the set of blowup points of uk consists of only onepoint p, and the location of p can be determined apriori by

5 γ(p) = 0.(1.8)

See Nagasaki and Suzuki [24], Li [19], Chen and Lin [10, 12] and the referencestherein. It is a fundamentally important question how to determine the signof ρk − 8π. In [11], the second and third authors have studied the questionfor the general case of multi-bubble solutions. For the case of single blowuppoint, the asymptotic formulas in [11] can be reduced to

ρk − 8π = c

[

(∆ log h(p)) + o(1)

]

εk log1

εk(1.9)

for some positive constant c > 0, where

εk = ρk

(

Ωh(x)euk(x)dx

)−1

,(1.10)

and uk is a solution of

∆uk + ρkh(x)euk(x)

h(x)euk(x)dx= 0 in Ω,

uk = 0 on ∂Ω.(1.11)

If the function log h is a harmonic function at p, then the formulas (1.9) doesnot yield any information for the sign of ρk − 8π. In this case, we shouldrefine our previous works in order to answer the questions here.

Theorem 1.3. Let Ω be a bounded piecewise C2 domain in R2 and h(x)be a C2 positive function in Ω. Suppose log h(x) is a harmonic function inΩ and uk is a sequence of blowup solutions of (1.11) with limh→+∞ ρk = 8π.Then

ρk − 8π =8

πh(p)

(

Ω

H(y, 0)

|y − p|4dy −∫

Ωc

dy

|y − p|4 + o(1)

)

εk,(1.12)

where p is the blowup point, Ωc = R2\Ω and H(y, 0) is defined by

H(y, 0) =h(y)

h(p)e8π(G(y,p)−γ(p)) − 1 for y ∈ Ω.(1.13)

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In many cases, it is convenient to work with the formulas (1.12). However,for some situations such as when the deformation of domains is required, itis suitable to work through the conformal mapping of the unit ball onto Ωwhile Ω is a simply connected domain. We explain it in the followings.

Let z = x1+ix2 denote the complex variable associated with the point x =(x1, x2), and z = f(w) be the conformal map from the unit ball w | |w| < 1onto Ω. Let w = g(z) stands for the inverse function of f . Obviously, the

Green function G(x, p) = − 12π

log |g(z)| and G(x, p) = − 12π

log( |g(z)||z−p|

). Hence,

γ(p) can be expressed in term of f by

γ(p) = G(p, p) =1

2πlog |f ′(0)| =

1

2πlog |a1|,(1.14)

where f(w) is written as

f(w) = f(0) +∞∑

n=1

anwn.(1.15)

Furthermore, due to the symmetry of G(x, y) = G(y, x), condition (1.8) isequivalent to 5xG(x, p) = 0 at x = p. Let ∂

∂zdenote the derivative with

respect to the complex variable z. Then at x = p,

∂zG(z, p)

z=p

= − 1

∂zlog

(

|g(z)|2|z − p|2

)∣

z=p

= − 1

g′(z)

g(z)− 1

z − p

z=p

= − 1

1

wf ′(w)− 1

f(w) − p

w=0

=a2

4πa21

.

Hence, (1.8) is equivalent to a2 = 0. Hence, for any critical point p of γ, weset

D(p) =∞∑

n=3

n2

n− 2|an|2 − |a1|2.(1.16)

Obviously, D(p) is well-defined because a2 = 0. Since the area |Ω| ofΩ = π

∑∞n=1 |an|2n, the series of (1.16) is absolutely convergent. By using

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the conformal map f , the asymptotic formulas (1.12) can be simplified to

Theorem 1.4. Suppose f and Ω be described as above and uk is a sequenceof blowup solution of (1.1) and (1.7). Let p be the blowup point of uk. Then

ρk − 8π = π(D(p) + o(1))εk,(1.17)

where εk and D(p) are given in (1.10) and (1.16).

We note that when Ω is smooth, the existence of solutions with singleone blowup point has been obtained by Weston and Moseley in 70s. In[33] and [21, 22], Weston and Moseley apply the Liouville theorem for (1.1)to construct approximation solutions, where the error of solutions and itsapproximations can be estimated within the range εlk for some large l. Thenthey use the implicit function theorem to obtain the solution with sufficientlysmall εk. For solutions with multi-bubbles, we refer the recent works Barakatand Pacard [3], Chen and Lin [12] and the references therein. However, inorder to construct solutions successfully, Weston and Moseley have to assumeΩ to satisfy a nondegenerate condition at p, that is, p is a nondegeneratecritical point of γ(x). In terms of f , the nondegenerate condition is equivalentto

|a3||a1|

6= 1

3.(1.18)

Under the condition (1.18), Theorem 1.4 has been proved by Suzuki andNagasaki [31] for those special solutions uk obtained by Weston and Moseley,because the estimates of the errors of solutions and its approximation is usedin their proof. Our theorem holds for any blowup solutions in any simply-connected domain. Our proof of Theorem 1.3 are based on the estimates ofthe error term of solutions and its approximation bubble, which has beendone for the general case in [11]. To see how D(p) is related to our problem,we employ the conformal map to construct a test function. Set vε(w) =2 log( 1+ε

ε+|w|2) for |w| ≤ 1 and uε(z) = vε(g(z)) for z ∈ Ω, where g is the

inverse function of f . Then by a straightforward computation (see section 5for details of computation), one has

J8π(uε) = 1 + 4πγ(p) + log|B1||Ω| + |a1|−2D(p)ε+O(|ε|2).(1.19)

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Clearly, if D(p) > 0 for some maximum point p of γ, then I8π(Ω) > 1 +

4πmaxΩ γ + log |B1||Ω|

. Hence by the ”if” part of Theorem 1.1, we concludethat Ω is of type C. In this paper, we will prove that the converse holds also.

Theorem 1.5. A bounded piecewise C2 simply-connected domain Ω is oftype C if and only if

D(p) > 0(1.20)

for some maximum point p of γ.

We use two deep results to prove Theorem 1.5. The first one is Theo-rem 1.3 or Theorem 1.4, that is, the asymptotic formulas (1.12) and (1.17).The other is the uniqueness theorem for solutions to equation (1.1) withρ ≤ 8π and for a simply connected domain Ω. By employing Bol’s isoperi-metric inequality, T. Suzuki [29] beautifully proved the uniqueness theoremfor ρ < 8π. Together with Theorem 1.3, this uniqueness theorem for ρ < 8πheuristically establishes that if D(p) < 0 for all maximum points of γ, thenΩ is not of type C. Because if D(p) < 0 for a maximum point p of γ, then by(1.18), p is a nondegenerate critical point of γ. Thus, there exists a sequenceof solution uk of (1.1) with ρ = ρk, which blows up at p. The existence ofsuch a sequence of solutions was due to Weston [33]. By Theorem 1.4, wehave ρk < 8π. Now, the uniqueness theorem of Suzuki yields that uk is themaximizer of Jρk

. Since the maximizer uk blows up at p as ρk ↑ 8π, it isplausible to guess that the maximizer of J8π does not exist, that is, Ω is notof type C. In fact, this is a corollary of the following stronger result.

Theorem 1.6. Suppose that Ω is a bounded piecewise C2 simple-connecteddomain, then equation (1.1) possesses at most one solution for each ρ ≤ 8π.Furthermore, the linearized equation of (1.1) with ρ ≤ 8π at the unique so-lution has a positive first eigenvalue.

As we noted before, Theorem 1.6 was proved by Suzuki [29] for ρ < 8π.For this paper, the uniqueness theorem for ρ = 8π plays an even more im-portant role. In fact, by the argument above, Theorem 1.6 for ρ = 8π impliesthat Ω is not of type C if D(p) < 0 for all maximum point p of γ. It is easy tosee that domains such as balls and cubes belong to this class. However, fordomains with D(p) = 0 for some maximum point p of γ, the above argument

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to prove that Ω is not type C simply fails, because Theorem 1.3 does not yieldany information of sign of ρk − 8π in this case. Surprisingly, the followingtheorem show that the existence of critical point p of γ with D(p) ≤ 0 set astrong restriction on the ”geometry” of Ω.

Theorem 1.7 Suppose that Ω is bounded piecewise C2 simply-connecteddomain and p is a critical point of γ with D(p) ≤ 0. Then p is the uniquemaximum point of γ. Furthermore, the unique solution uρ(x) of (1.1) withρ < 8π must blow up at p as ρ ↑ 8π.

Together with Theorem 1.5, we have

Theorem 1.8. Suppose that Ω is a bounded piecewise C2 simply-connecteddomain and γ(x) has more then one maximum point. Then Ω is of type C.

The paper is organized as follows. In order to prove (1.5), we have toshow apriori that for a sequence of blowup solution, any blowup point mustbe inside of the domain Ω. This has been done for C2 domain. This holdsalso for a peiecewisely C2 domain. In section 2, we give a grief account. Insection 3, we first prove Theorem 1.3. Here we have to use estimates in [11]for the error term ηk of the solution uk and its approximation. By using thisestimate, we will derive the asympototic formulas (1.12). This asymptoticformulas is more useful when we come to computation for concrete examples.By using the conformal map f , (1.17) can be derived easily from (1.12). Aswe note, Theorem 1.6, the uniqueness theorem, plays an important role in ourpaper. We present its proof in section 4. Here, we employ the symmetrizationand the Bol inequality to reduce our problem to the case of radial functions.We treat the radial case in a more elementary and direct way than in [29],where the uniqueness theorem for ρ < 8π was proved in smooth domains.Our argument becomes lengthy when non-smooth domains are considered.However, we want to include its proof here because the uniqueness theoremfor non-smooth domains has its own interest. As we mentioned, the asymp-totic formulas (1.17) is employed in the argument when the deformation ofdomains is required. This is the case for the proof of Theorem 1.7, whoseproof are given in section 5. In section 6, we first give a proof of Theorem1.5 and by using Theorem 1.5, we prove Theorem 1.1. In the final section,we present several examples to discuss the type of domains, as applications

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of our theorems in this paper.

Ackowledgement. Part of this work has been done while the second au-thor visited National Center for Theoretical Sciences of NSC in Taiwan. Hewould like to thank NCTS for the warm hospitality and the stimulating en-vironment.

2 Non-smooth domains

Since in section 6, we will consider non-smooth domains such as n-polygons.We will give a brief account concerning the regularity of solutions of (1.1) inour setting. First, we note that in order to derive (1.5) as done in the proofof Theorem 7.1 in [5], we have to prove the blowup point could not occur onthe boundary. This is known when Ω is a C2 domain. For a piecewise C2

domain, this holds true as shown in the followings.

Lemma 2.1. Let u ∈ H10 (Ω) be a weak solution of (1.1). Then u ∈

C∞(Ω)∩C(Ω). Furthermore if uk is a sequence of blowup solutions of (1.1),then all the blowup points are located inside of Ω.

Proof. By the Moser-Trudinger inequality, eu(x) ∈ Lp(Ω) for any p > 1.Thus, u(x) ∈ C∞(Ω) follows readily from the standard regularity theoremsof elliptic equation. As for the boundary regularity, u is C1,α locally nearx0 ∈ ∂Ω for any α < 1 provided ∂Ω has a C2 regularity at x0. Now supposex0 is one of vertex Q1, . . . , QN, say, x0 = 0 ∈ Γ1 ∩ Γ2 and θ is the innerangle of Γ1 and Γ2 at 0, where Γj is the C2 connected component of ∂Ω.Let w = g(z) be the conformal map which maps Γ1 ∪ Γ2 to be a C2 curvenear g(0). Set g(0) = 0 and z = f(w) be to be the inverse map of g. Thenv(w) = u(f(w)) satisfies

∆v + |f ′(w)|2ev = 0 in Ω,v(w) = 0 for w ∈ ∂ ′Ω = g(Γ1 ∪ Γ2),

(2.1)

where Ω = g(Bδ0(0) ∩ Ω). Clearly, |f(w)| = |w| θπ (1 + o(1)) with o(1) → 0

as |w| → 0. Hence, if θ > π, then |f ′(w)|2 is a Holder function of exponentα0 for some α0 > 0. By the regularity theorem, v ∈ C1,α near 0 for anyo < α < 1. In particular, u(x) is continuous at 0.

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If 0 < θ < π, then |f ′(w)|2ev ∈ Lp0 for some p0 > 1 due to the Moser-Trudinger inequality. Since ∂Ω is C2 at 0, v ∈ W 2,p0 locally near 0. By theSobolev embedding, v is Holder locally near 0. Hence the continuity of u ateach vortex is proved.

Now suppose the exists a sequence of solutions uk of (1.1) with ρk → ρ.It suffices for us to show that uk(x) can not blow up at any vertex pointQ. If the angle θ at Q is less than π, then the domain Ω is converx in aneighborhood of 0. Applying the method of moving planes to Ω, the blowuppoints must be away from Q. If θ > π, then we can use the conformal map

f(w) = wθ

θ1 , Q = 0 and f(0) = 0 is assumed where θ1 < π. Clearly, f(w)map a curve C onto Γ1 ∪Γ2 and the inner angle of C at 0 is equal to θ1. LetΩ be the domain such that f(Ω) = Ω ∩ Bδ0(0). Then Ω is convex at 0, andv(w) = u(f(w)) satisfies

∆v + |w|2(θθ1

−1)ev = 0 in Ω,

v = 0 on ∂′Ω = g(Γ1 ∪ Γ2),

where g = f−1 is the inverse function. Let w = (w1, w2). By a rotation,we may assume the axis w1 = 0 intersects with ∂ ′Ω non-tangentially and

Ω ⊂ w | w2 < 0. Since θθ1−1 > 0, |w|2(

θθ1

−1)is increasing in |w2| for w ∈ Ω,

Thus, we can apply the method of moving plane to show v(w) > v(wλ) forw ∈ Ω, w2 < λ < 0 and wλ is the reflection point of w with respect tow2 = λ, provided that Ω ∩ w | w2 = λ is non-empty. Then it implies thatthe blowup points must be away from Q. This completes the proof of Lemma2.1. Q.E.D.

Corollary 2.2. Suppose that Ω is a bounded piecewise C2 domain. If Ω isnot type C, then

I8π(Ω) = 1 + 4π supΩ

γ(x) + log|B1||Ω| .(2.2)

Proof. This immediately follows from the original proof of (2.2) for thesmooth domains and Lemma 2.1, which states the blowup point must be in-side of Ω. We skip the proof here and refer the detail of the proof to Theorem7.1 of [5]. Q.E.D.

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3 Asymptotic expansion of ρk − 8π

In this section, we are going to prove the asymptotic formulas (1.12) fora piecewisely C2 doamin Ω. Since all the estimates of solutions are local innature, for the simplicity, we may assume our domain Ω is C2 throughoutthis section. Let Ω be a C2 bounded domain and h(x) be a positive C2

function defined in Ω. Consider a sequence of blowup solution uk of (1.11)such that

limk→+∞

ρk = 8π.(3.1)

Set

λk = uk(pk) = maxΩ uk(x) → +∞ as k → +∞,

p = limk→+∞ pk.(3.2)

To describe the behavior of uk, we collect some well-known facts in thefollowing lemmas.

Lemma 3.1. By passing to a subsequence, solutions uk(x) converges tothe Green function G(x, p) in C2

loc(Ω\p). Furthermore, the blowup point psatisfies

5 (log h(x) + 8πG(x, p)) = 0 at x = p.(3.3)

Lemma 3.1 was proved by Nagasaki and Suzuki [24] and the identity (3.3)can be derived from the Pohozaev identity. Since the proofs are standardand the results are well-known now, we skip the proofs here. To describe thebubbling behavior of uk near p, we have to quote a result due to Li [19]. Wedenote uk by

uk = uk(x) − dk, dk = log

(

Ωh(x)eukdx

)

.(3.4)

Then uk(x) satisfies

∆uk + ρkh(x)euk(x) = 0 in Ω,

uk(x) = −dk on ∂Ω.(3.5)

Since uk(x) uniformly converges to G(x, p) for |x − pk| = δ0 for any fixedsmall δ0 > 0, the difference |uk(x)− uk(y)| is uniformly bounded for any two

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points x, y on ∂Bδ0(pk). Thus, we can apply Li’s theorem to uk. Hereafter,Bδ0(p) denote the ball of center p and radius δ0. In [19], Li proved

Lemma 3.2. There exists constants c and δ0 such that

|uk(x) − λk + 2 log(1 + ekeλk |x− pk|2)| ≤ c(3.6)

for |x− pk| ≤ δ0, where λk = maxΩ uk = λk − dk and ek = ρkh(pk)8

.

For |x− pk| = δ0, (3.6) implies

|uk(x) + λk| ≤ c.(3.7)

Since uk(x) uniformly converges to G(x, p) for |x−pk| ≥ δ0, we have |uk(x)+dk| = |uk(x)| ≤ c1 for |x− pk| ≥ δ0. Thus by (3.7),

|dk − λk| ≤ c1 and |λk − 2λk| ≤ c1.(3.8)

As in [11], the estimate (3.6) is not enough for us to derive (1.12). There-fore, we have to estimate the profile of uk more precisely in Bδ0(p). In orderto achieve it, we introduce the error term ηk by

ηk(x) = uk(x) − Uk(x) − ρk(Gk(x, pk) − γ(pk)),(3.9)

for x ∈ Ω, where

Uk(x) = λk − 2 log

[

1 + ekeλk |x− qk|2

]

,(3.10)

and qk satisfies5 Uk(pk) = 5 log h(pk).(3.11)

Note that 5Uk(x) = −4ekeλk (x−qk)

1+ekeλk |x−qk|2

. From it, qk satisfies

|pk − qk| ≤ c e−λk .(3.12)

Clearly from (3.3), we expect that 5ηk(pk) = −5Uk(pk)− ρk 5 G(pk, pk) =−[5 log h(pk) + ρk 5 G(pk, pk)] should be small. In fact, we have

| 5 ηk(pk)| = O(λke−λk).(3.13)

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See Lemma 5.4 of [11]. Recall that Uk(x) is an entire solution of

∆Uk + ρkh(pk)eUk(x) = 0 in R2.

For the rest of this section, we denote Gk(x) = G(x, pk) and Gk(x) = G(x, pk)to simplify our notation. First, we derive the equation for ηk.

Lemma 3.3. The error term ηk satisfies

∆ηk(y) + ρkh(pk)eUk(y)Hk(y, ηk) = 0 for y ∈ Ω,

ηk(y) = O(λke−λk) on ∂Ω,

(3.14)

where

Hk(y, t) =h(y)

h(pk)et+ρk(Gk(y)−Gk(pk)) − 1.

Proof. (3.14) is already proved in [11]. See Theorem 1.4 of [11]. For theconvenience of readers, we give a scatch of the proof. Since Gk(x) is harmonic,(3.9) yields

∆ηk = ρkh(pk)eUk(x) − ρkh(x)e

uk

= −ρkh(pk)eUk(x)

h(x)

h(pk)euk−Uk(x) − 1

= −ρkh(pk)eUk(x)

h(x)

h(pk)eηk+ρk(Gk(x)−Gk(pk)) − 1

by applying uk − Uk(x) = ηk + ρk(Gk(x) − Gk(pk)).For y ∈ ∂Ω, we have

ηk(y) = −dk − Uk(y) +ρk

2πlog |y − pk| + ρkGk(pk)

= −dk + ρkGk(pk) + λk + 2 log

(

ρkh(pk)

8

)

+ρk

2πlog

(

|y − pk||y − qk|

)

+2 log

(

1 +8

ρkh(pk)|y − qk|−2e−λk

)

+O(|ρk − 8π|)

= −dk + ρkGk(pk) + λk + 2 log

(

ρkh(pk)

8

)

+O(λke−λk).

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Here, we use (3.12) and |ρk − 8π| = O(λke−λk). For the proof of the latter,

see Theorem 1.1 of [11]. In [11], we denote

Ik = −dk + λk + ρkGk(pk) + 2 log

(

ρkh(pk)

8

)

.

By using Theorem 7.2 in [11],

|Ik| ≤ c λke−λk .(3.15)

Thus, ηk(y) = O(λke−λk) for y ∈ ∂Ω. Q.E.D.

The estimate (3.15) yields the following.

Lemma 3.4. e−λk = π2h2(p)e8πγ(p)e−dk [1 + o(1)].

Proof of Theorem 1.3. By the definition of Hk(x, t), we have

ρkh(x)euk(x) = ρkh(pk)e

Uk(x) + ρkh(pk)eUk(x)Hk(x, ηk(x))(3.16)

for x ∈ Ω. Substituting (3.16) into (1.11),

ρk =∫

Ωρkhe

uk(x)dx(3.17)

=∫

Ωρkh(pk)e

Uk(x)dx+∫

Ωρkh(pk)e

Uk(y)Hk(y, ηk)dy

= 8π −∫

Ωcρkh(pk)e

Uk(y)dy +∫

Ωρkh(pk)e

Uk(y)Hk(y, ηk)dy.

Clearly

Ωcρhh(pk)e

Uk(y)dy(3.18)

=

(

ρkh(pk)

e2k

Ωc

dy

|y − pk|4+ o(1)

)

e−λk

=

(

8

πh(p)

Ω

dy

|y − p|4 + o(1)

)

e−λk .

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The integrand of the last term of (3.17) can be expressed, by the Taylorexpansion, as

ρkh(pk)eUk(y)Hk(y, ηk)(3.19)

= ρkh(pk)eUk(y)Hk(y, 0) + ρkh(pk)e

Uk(y)(Hk(y, 0) + 1)ηk

+ρkh(pk)eUk(y)O(|ηk|2).

Except for the first item of (3.19), heuristically the integration of all otherterms in (3.19) should bounded by o(εk). However, this o(εk)-estimate of theintegral of ρkh(pk)e

Uk(y)ηk is not obvious at all. In fact, its smallness is dueto the effect of cancellation. This effect of cancellation can be made clearonly when ηk is explicitly solved in the order of εk. To avoid the complexitynecessary for this delicate estimates, we apply an useful trick from [11]. Set

ψ(x) =1 − a|x− qk|21 + a|x− qk|2

,(3.20)

where a = ρkh(pk)8

eλk . The function ψ(x) is chosen as a comparison functionbecause it satisfies

∆ψ(x) + ρkh(pk)eUk(y)ψ = 0 in R2.(3.21)

Since ηk satisfies

∆ηk + ρkh(pk)eUk(y)Hk(y, ηk) = 0 in Ω.

by Lemma 3.3, we have

Ωρkh(pk)e

UkHk(y, ηk)dy = −∫

∂Ω

∂ηk

∂νdσ(3.22)

=∫

∂Ω

(

ψ∂ηk

∂ν− ηk

∂ψ

∂ν

)

dσ +∫

∂Ω

(

ηk∂ψ

∂ν− (1 + ψ)

∂ηk

∂ν

)

dσ.

Applying (3.22) and (3.21) together, the Green Theorem yields

∂Ω

(

ψ∂ηk

∂ν− ηk

∂ψ

∂ν

)

dσ = −∫

Ωρkh(pk)e

Uk(y)(Hk(y, ηk) − ηk)ψdy.

Note that

ψ = −1 +2

1 + a|x− qk|2.

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Ωρkh(pk)e

Uk(y)Hk(y, 0)(−ψ)dy

=∫

Ωρkh(pk)e

Uk(y)Hk(y, 0)dy − 2∫

Ω

ρkh(pk)eUk(y)

(1 + a|x− qk|2)Hk(y, 0)dy.

By Lemma 3.3,

h(pk)Hk(y, 0)(3.23)

= explog(h) + ρk(Gk(y) − Gk(pk)) − h(pk)

=2∑

j=1

aj(yj − pk,j) +B11

2(y1 − pk,1)

2 +B22

2(y2 − pk,2)

2

+1

2

(

2∑

j=1

aj(yj − pk,j)

)2

+O(|y − pk|3),

where a = (a1, a2) = 5[log h(y) + ρkGk(y)] at y = pk, and

(

B11 00 B22

)

is

the Hessian of log h+ ρkGk at y = pk, after orthogonal transformation of thecoordinate. By (3.13), we have

|a| = | 5 log h(y) + ρkGk(y)| ≤ λke−λk

for y = pk. Since |pk − qk| = O(e−λk) by (3.12), (3.23) yields

h(pk)Hk(y, 0) =2∑

j=1

bj(yj − qk,j) +1

2

2∑

j=1

Bjj(yj − qk,j)2(3.24)

+O(λke−2λk + λ2

ke−2λk |y − qk|2 + |y − qk|3).

NoteB11 +B22 = 0.

because log h is harmonic. Recall that Uk(y) is radially symmetric withrespect to qk. Thus,

ρk

Bδ(qk)

eUk(y)

2∑

j=1

bj(yj − qk,j) +1

2

2∑

j=1

Bjj(yj − qk,j)2

dy = 0.

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Therefore, by (3.24) we have∣

ρk

Bδ(qk)h(pk)e

Uk(y)Hk(y, 0)dy

(3.25)

≤ c

Bδ(qk)

eUk(y)|y − qk|3dy + λke−2λk

≤ c

δe−λk + λke−2λk

,

where c is a constant independent of δ and k.For any arbitrary small δ > 0,

ρk

Ω\Bδ(qk)h(pk)e

Uk(y)Hk(y, 0)dy

= ρke−2k e−λkh(pk)

Ω\Bδ(qk)

Hk(y, 0)

|y − qk|4dy +O(1)e−2λk

Ω\Bδ(qk)

Hk(y, 0)

|y − qk|6dy

=8

πh(p)e−λk

(

Ω\Bδ(p)

H(y, 0)

|y − p|4dy + o(1)

)

,

because|eUk(y) − e−2

k |y − qk|−4e−λk | ≤ c |y − qk|−6e−2λk

for |y − qk| >> e−12λk . With (3.25), it yields

ρk

Ωh(pk)e

Uk(y)Hk(y, 0)dy =8

πh(p)e−λk

(

Ω

H(y, 0)

|y − p|4dy + o(1)

)

,(3.26)

where∫

Ω

H(y, 0)

|y − p|4dy := limδ↓0

Ω\Bδ(p)

H(y, 0)

|y − p|4dy.

By the scaling y′ = eλk2 y, it is easy to obtain

Ω

eUk(y)

1 + a|y − qk|2Hk(y, 0)dy

= O(e−32λk)(3.27)

by (3.24). By Lemma 3.3, |ηk(y)|+ |5ηk(y)| = O(λke−λk) for y ∈ ∂Ω. Hence

∂Ω

(

ηk∂ψ

∂ν− (1 + ψ)

∂ηk

∂ν

)

dσ = O(λke−2λk),(3.28)

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and (1.12) follows readily from (3.18), (3.26) – (3.28). This proves Theorem1.3. Q.E.D.

Proof of Theorem 1.4. To derive (1.17), we use a conformal mappingz = f(w) from the unit ball B1 → Ω such that

f(0) = p.

Since p is a critical point of γ, we have

f ′′(0) = 0.

To simplify our notation, we denote the function uk(f(w)) by uk(w). Thenuk satisfies

∆uk + ρkh(w)euk

B1

h(w)eukdw= 0 in B1,

uk = 0 on ∂B1,

where h(w) = |f ′(w)|2 and dw stands for the volume form in B1. By thechoice of f , the blowup point of uk is the origin and Gk(x) → 0 uniformly inB1. Thus

h(0)H(w, 0) = |f ′(w)|2 − |f ′(0)|2, and h(0) = |f ′(0)|2 = |a1|2.(3.29)

Let f(w) = f(0) +∑∞n=1 anw

n. Then (3.29) implies

h(0)∫

B1

H(w, 0)

|w|4 dw = 2π∫ 1

0r−3

∞∑

n=3

|an|2n2r2(n−1)dr(3.30)

= 2π∑

n=3

|an|2n2∫ 1

0r2n−5dr

= π∞∑

n=3

n2

n− 2|an|2.

On the other hand,∫

Bc1

dy

|y|4 = 2π∫ ∞

1

dr

r3= π.

Therefore, we have by (1.12)

ρk − 8π =8

h2(0)e−λk

(

∞∑

n=3

n2

n− 2|an|2 − |a1|2

)

.(3.31)

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By Lemma 3.4, we have

e−dk =1

h2(0)π2e−λk(1 + o(1)),

and

εk = ρke−dk =

8

πh(0)2e−λk .(3.32)

Substituting (3.32) into (3.31), we have (1.17). Q.E.D.

4 Uniqueness Theorem

The main purpose of this section is to prove Theorem 1.6. First, we startwith the beautiful theorem of uniqueness of solution (1.1) with ρ < 8π, whichwas due to T. Suzuki [29].

Theorem 4.1. Suppose that Ω is a bounded piecewise C2 simply-connecteddomain. Then for each ρ ∈ [0, 8π), there exists an unique solution for (1.1).

By applying the classical Bol inequality, Suzuki [29] proved Theorem 4.1for smooth domains by virture of a special technique of symmetrization,which was initiated by Bandle [2] first. Without employing Bol’s inequality,this method of symmetrization has been successfully applied to other prob-lems related to equation (1.1) in recent years, e.g., see [9], [13]. Our proof ofTheorem 1.6 also employs the method of symmetrization, which could allowthe original problem to be reduced to the radial case. We treat the radialcase in more elementary and straightforward way than in [29] for ρ < 8π.However, our complete proof becomes lengthy because non-smooth domainsare considered here. We would like to give a proof for non-smooth domain,because it has its own interest. We begin with the following isoperimetricinequality, which was due to Bandle [2] when ω is a simply-connected domain.

Lemma 4.2. Suppose v satisfies ∆v + ev = 0 in Ω and satisfies

Ωev(x)dx ≤ 8π,(4.1)

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where Ω is a simply-connected domain. Set

m(ω) =∫

ωevdx and l(∂ω) =

∂ωe

v2 dσ

for any ω ⊂⊂ Ω. Then the inequality holds

2l2(∂ω) ≥ [(8π −m(ω)]m(ω).(4.2)

Furthermore, if ω is not simply-connected, then the inequality (4.2) is alwaysstrict.

Proof. When ω is simply-connected, (4.2) was proved by Bandle [2]. Forthe general case, the proof can be given by induction on the number of ”holes”of ω. For the simplicity, we assume ∂ω = Γ1 + Γ2 and Γ2 bounds anothersimply connected domain Ω. Then by (4.2) for simply-connected domains,we have

2l2(∂ω) = 2[l(Γ1) + l(Γ2)]2

= 2l2(Γ1) + 2l2(Γ2) + 4l(Γ1)l(Γ2)

> [8π −m(ω ∪ Ω)]m(ω ∪ Ω) + [8π −m(Ω)]m(Ω)

= [8π −m(ω) −m(Ω)][m(ω) +m(Ω)] + [8π −m(Ω)]m(Ω)

= [8π −m(ω)]m(ω) + 2m(Ω)[8π −m(Ω) −m(ω)]

≥ [8π −m(ω)]m(ω),

because m(Ω) +m(ω) ≤ ∫

Ω evdx ≤ 8π. Q.E.D.

Let Ω be a piecewisely C2 simply-connected domain, and ∂Ω =⋃Nl=1 Γl,

where Γl is a connected component of C2 portion of the boundary ∂Ω.

Lemma 4.3. Let µ2(Ω) be the second eigenvalue of ∆+ev for the Dirichletproblem. Then

µ2(Ω) > 0,

provided that∫

Ω evdx ≤ 8π.

Proof. We prove Lemma 4.3 by contradiction. Suppose µ2(Ω) ≤ 0. Thenthere exists a positive constant K ≤ 1 and a second eigenfunciton ϕ satisfying

∆ϕ +Kevϕ = 0 in Ω,ϕ = 0 on ∂Ω.

(4.3)

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Before continuing to give a proof, we want to make a remark concerning thecontinuity of ϕ. It is natural that ϕ is assumed to be in H1

0 (Ω). By composingwith a conformal map as done in Lemma 2.1, we can prove ϕ ∈ C∞(Ω)∩C(Ω).Since it is standard, we skip the proof here.

Let Ω+ = x ∈ Ω | ϕ(x) > 0. Set U(x) = −2 log(1 + 18|x|2) which

satisfies ∆U + eU = 0 in R2. Without loss of generality, we may assume∫

Ω+ev(x)dx ≤ 4π,(4.4)

because the total measure∫

Ω ev(x)dx ≤ 8π. For any t > 0, the set Ωt = x ∈

Ω|ϕ(x) > t is compactly contained in Ω because ϕ is continuous on Ω. Weset r(t) ≥ 0 such that

Br(t)

eU(x)dx =∫

ϕ>tev(x)dx,(4.5)

where Br(t) is the open ball of center 0 and radius r(t). Note that by theregularity of ϕ, for any t < s < maxΩ ϕ, the set x | t ≤ ϕ(x) < s has apositive measure. Therefore r(t) is strictly decreasing in t and is continuousin t for t ∈ (0,maxΩ ϕ). Denote ϕ∗(r) to be the symmetrization of ϕ withrespect to the measures eU(x)dx and ev(x)dx respectively, that is,

ϕ∗(r) = supt | r < r(t).(4.6)

Then ϕ∗(x) has the same distribution of ϕ with respect to measures eU(x)dx

and ev(x)dx respectively, that is, for t > 0,∫

ϕ∗>teU(x)dx =

ϕ>tev(x)dx.(4.7)

Therefore,∫

BR0

eU(x)(ϕ∗)2dx =∫

Ω+ev(x)ϕ2dx(4.8)

holds where R0 = r(0+) = limt↓0 r(t). To derive a contradiction, we use thecoarea formulas

− d

dt

Ωt| 5 ϕ|2dx =

∂Ωt

| 5 ϕ|ds, and(4.9)

− d

dt

Ωt

ev(x)dx =∫

∂Ωt

ev

| 5 ϕ|ds

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hold for almost everywhere t. Since∫

Ω evdx ≤ 8π, we have by the co-area

formulas (4.9) and by Lemma 4.2,

− d

dt

Ωt

| 5 ϕ|2dx =∫

ϕ=t| 5 ϕ|ds(4.10)

≥(

ϕ=te

v2 ds

)2(∫

ϕ=t

ev

| 5 ϕ|ds)−1

= −[

d

dt

(∫

Ωt

ev(x)dx

)

]−1

l2(ϕ = t)

≥ 1

2

(

8π −∫

Ωt

ev(t)dx

)(∫

Ωt

ev(x)dx

)

[

− d

dt

Ωt

ev(x)dx

]−1

=1

2

(

8π −∫

Ω∗

t

eU(x)dx

)(

Ω∗

t

eU(x)dx

)[

− d

dt

Ω∗

t

eU(x)dx

]−1

,

for almost everywhere t where Ω∗t = Br(t) = x | ϕ∗(x) > t. Since U(r)

is a radial function, (4.2) becomes an equality for any ball Br(t). Thus, thecoarea formulas (4.9) yields

− d

dt

(

Ω∗

t

| 5 ϕ∗|2dx)

(4.11)

=1

2

(

8π −∫

Ω∗

t

eU(x)dx

)(

Ωtt

eU(x)dx

)[

− d

dt

Ω∗

t

eU(x)dx

]−1

.

Together with (4.10), (4.11) implies

− d

dt

Ωt

| 5 ϕ|2dx ≥ − d

dt

Ω∗

t

| 5 ϕ∗|2dx(4.12)

for almost everywhere t. By integrating (4.12) with respect to t, we obtainthen

BR0

| 5 ϕ∗|2dx ≤∫

Ω+

| 5 ϕ|2dx.(4.13)

Together with (4.8) and (4.13), the equation (4.3) yields

0 ≥ (K − 1)∫

Ω+ev(x)ϕ2(x)dx(4.14)

=∫

Ω+| 5 ϕ|2dx−

Ω+ev(x)ϕ2(x)dx

≥∫

BR0

| 5 ϕ∗|2dx−∫

BR0

eU(x)|ϕ∗|2(x)dx.

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Therefore, the first eigenvalue λ1 of ∆ + eU on BR0 is nonpositive.Since U(x) = U(|x|) is radial, the first eigenfunction of ∆ + eU(r) is also

radial. By a straightforward computation, the function ψ(r) = 8−r2

8+r2satisfies

∆ψ(r) + eU(r)ψ(r) = 0 in R2.(4.15)

Therefore, the nonpositive first eigenvalue λ1 of ∆ + eU implies R0 ≥√

8.On the other hand, since

BR0

eU(x)dx =∫

Ω+ev(x)dx ≤ 4π,

we deduce R0 =√

8, λ1 = 0 and∫

BR0

eUdx = 4π.(4.16)

Consequently, the inequality of (4.12) and then all the inequalities of (4.10),turn out to be equalities. Particularly, by Lemma 4.2, Ωt is simply connectedfor almost every t. Due to the regularity of ϕ, Ωt and then Ω+ are all simply-connected domains for all t > 0. Similarly, the Schwartz inequality on eachlevel set ϕ = t becomes an equality and it implies

ev(x) = Φ+(ϕ(x))| 5 ϕ(x)|2(4.17)

holds for almost everywhere t = ϕ(x) and for some function Φ+ of t.By (4.16), we have

Ω−

ev(x)dx ≤ 4π,

where Ω− = x | ϕ(x) < 0. By the same argument of symmetrization, wehave ∫

Ω−

ev(x)dx = 4π,(4.18)

Ω− is also a simply connected domain and

ev(x) = Φ−(ϕ(x))| 5 ϕ(x)|2(4.19)

for almost everywhere s = ϕ(x) < 0 and a function Φ− of s.Since both Ω+ and Ω− are simply connected, the nodal line, that is, the

closure of ϕ(x) = 0 | x ∈ Ω, must intersect with ∂Ω and at least, one of

23

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∂Ω± ∩ ∂Ω contains an arc of positive length. Without loss of generality, onemay assume ∂Ω+∩∂Ω contains an arc of positive length. Furthermore, for anytwo points yk ∈ ∂Ω∩∂Ω+, k = 1, 2, which are not on the nodal line, we choosea sequence of xi,k ∈ Ω → yk for k = 1, 2 such that ϕ(xi,k) = ti ↓ 0 where(4.17) holds for such ti. Since Φ+(ϕ(xi,1)) = Φ+(ϕ(xi,2)) and ev(xi,k) → 1 asi→ +∞ for k = 1, 2, (4.17) implies | 5 ϕ(y1)| = | 5 ϕ(y2)|, i.e., | 5 ϕ(y)| =constant for all y which are contained in a connected component of Γl∩∂Ω+,l = 1, 2, . . . , N . We recall Γl is a C2 portion of the boundary ∂Ω.

Since the nodal line has a non-empty intersection with ∂Ω, we let x0

denote any point of the intersection (Note that the nodal line is continuousup to the boundary because it has a finite length.) Two cases are discussedseperately.Cases 1. Suppose x0 ∈ Γl for some l ∈ 1, 2, . . . , N. Then by the Lp

elliptic regularity, ϕ(x) is W 2,p in a neighborhood of x0 with any p > 1.Thus ϕ(x) ∈ C1,α at x0 for any α ∈ (0, 1). Since x0 is the intersection of theboundary and the nodal line of ϕ, we have 5ϕ(x0) = 0. Then | 5 ϕ(y)| = 0for all y ∈ Γl∩∂Ω+, which yields a contradiction to the Hopf boundary pointlemma, unless ϕ(x) is identically to be zero. This proves Lemma 4.3 for case1.Cases 2. Suppose x0 is not a smooth point of the boundary, say, x0 ∈Γ1 ∩ Γ2. Let θ denote the inner angle of Γ1 and Γ2 at x0. For the simplicityof notations, we assume x0 = 0. By case 1, we conclude that one of Γ1 andΓ2 must be contained in ∂Ω+. Without loss of generality, we may assumeΓ1 ⊆ ∂Ω+. Let w = g(z) be a conformal mapping which maps Ω∩Bδ0(0) intoΩ such that one part of the boundary Ω, which is the image of ∂Ω ∩ Bδ0(0)under g, is C2 at the image of g(0). Let g(0) = 0 and f(w) denote the inversemap of g. Set v(w) = v(f(w)) and ϕ(w) = ϕ(f(w)). Then ϕ satisfies

∆ϕ + |f ′(w)|2eϕv = 0 in Ω,ϕ = 0 in ∂′Ω,

(4.20)

where ∂′Ω = ∂Ω ∩ g(∂Ω ∩Bδ0). Since θ is the inner angle of Γ1 and Γ2 at 0,we have

|f ′(w)|2 = |w|2( θπ−1)(1 + o(1)),(4.21)

where o(1) → 0 as |w| → 0.If π < θ < 2π, then |f ′(w)|2 is a Cα0 function for some α0 > 0. By the

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regularity of elliptic equation, ϕ is C1,α near 0 for any α ∈ (0, 1). Thus,

5 ϕ(0) = 0 and | 5 ϕ(w)| ≤ cα|w|α(4.22)

for any 0 < α < 1 because the nodal line of ϕ touch ∂ ′Ω at 0. On the otherhand, for w = w(z), z ∈ Γ1, we have

| 5w ϕ(w)| = | 5z ϕ(z)|| ∂z∂w

| = c |f ′(w)|,(4.23)

where | 5 ϕ(z)| = c for all z ∈ Γ1. Choose θπ− 1 < α < 1 and w = w(z)

where z ∈ Γ1 ⊆ ∂Ω+. Then by (4.22) and (4.21), we have

c |w| θπ−1 ≤ cα|w|α,

which implies c = 0. Hence, we obtain a contradition again.Suppose 0 < θ < π. Then ϕ isH1 in a neighborhood of 0. Since |f ′(w)|2 =

O(|w| θπ−1), we have |f ′(w)|2evϕ ∈ Lp locally near 0 for any 1 < p < 1

1− θπ

.

By the elliptic regularity and Sobolev’s embedding theorem, ϕ ∈ Cα0 locallyfor some α0 < 1. We want to prove ϕ ∈ Cα locally for any α ∈ (0, 1) byiteration. To see it, we assume ϕ ∈ Cα0 . Then

|ϕ(w)| ≤ c0|w|α0.(4.24)

Substituting (4.24) into the equation, we have

|f ′(w)|2ev|ϕ(w)| ≤ c1|w|(2θπ

+α0−2).

By the elliptic regularity, ϕ ∈ W 2,p locally near 0 for

1 < p <

(

2

2 − (2θπ

+ α0)

)

+

=

+∞ if 2 ≤ 2θπ

+ α0,2

2−( 2θπ

+α0)if 2 > 2θ

π+ α0.

If 2 ≤ 2θπ

+ α0, then ϕ ∈ Cα locally near 0 for any α ∈ (0, 1). So, we assume2 > 2θ

π+ α0. Then by the Sobolev embedding, ϕ ∈ Cα for any α ∈ (0, 1) if

2θπ

+α0 > 1, and ϕ ∈ Cα for any α < 2θπ

+α0 if 2θπ

+α0 ≤ 1. In particular, ϕ ∈Cα0+ θ

π . By repeating the process a finite time, we have established ϕ ∈ Cα

locally near 0 for any α. Then |ϕ(w)| ≤ cα|w|α for α ∈ (0, 1). Substitutingit into equation (4.20) again, by noting |f ′(w)|2evϕ(w) = O(|w|−β) for some

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β < 1, we have |f ′(w)|2evϕ(w) ∈ Lp locally for some p > 2. Applying theregularity theorem and the Sobolev embedding once again, we conclude thatϕ ∈ Cα1 locally near 0 for some α1 > 0.

Since the nodal line of ϕ touch the boundary at 0, 5ϕ(0) = 0. In partic-ular, for w = w(z), 0 6= z ∈ Γ1 and z → 0,

0 = | 5 ϕ(0)| = limw→0

| 5 ϕ(w)|

= limz→0

| 5 ϕ(z)||w| θπ−1(1 + o(1)).

Recall that |5ϕ(z)| = constant for z ∈ Γ1. By the identity above, we deduce| 5 ϕ(z)| = 0 for z ∈ Γ1, which yields a contradiction to the Hopf boundarypoint lemma. This contradiction finishes the proof of Lemma 4.3 Q.E.D.

Remark 4.4. If∫

Ω evdx < 8π, then a contradiction can be obtained by

using (4.16) and (4.18) only, that is, no assumption of smoothness of ∂Ω isrequired for Lemma 4.3 when ρ < 8π. However, in order to apply the sym-metrization, the continuity of ϕ on Ω is needed. For example, this continuityof ϕ is guaranteed if Ω is a Lipschitz domain. Even for a domain such asΩ0 = z | z = 1

3w3+w, |w| > 1, this holds true. Note that ∂Ω0 has a cusp at

±23i and the inner angles at ± 2

3i are equal to 2π. Hence Theorem 4.1 holds for

Ω0. However, for ρ = 8π, it is still an open problem, because the nodal line ofϕ might touch with the boundary ∂Ω0 at ±2

3i, where our method simply fails.

Proof of Theorem 1.6. We first prove that the linearized equation of(1.1) at u has no null eigenfunctions in H1

0 (Ω). Suppose ϕ is a solution ofthe linearized equation

∆ϕ+ρeuϕ∫

Ω eudx

− ρ(∫

Ω euϕdx)eu

(∫

Ω eudx)2

= 0 in Ω,

ϕ = 0 on ∂Ω,

(4.25)

where ρ = 8π and u is a solution of (1.1) with ρ = 8π. For ρ < 8π, we referethe proof to [29].

By the remark in the proof of Lemma 4.3, ϕ is continuous on Ω. We wantto prove ϕ ≡ 0 in Ω. Let

ϕ = ϕ−∫

Ω euϕdx

Ω eudx

, and v = u+ log ρ− log(∫

Ωeudx)(4.26)

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Then ϕ satisfies

∆ϕ+ evϕ = 0 in Ω,ϕ |∂Ω= constant on ∂Ω.

(4.27)

By (4.26), ϕ also satisfies∫

Ωevϕ(x)dx = 0(4.28)

By (4.28), ϕ must change sign provided that ϕ 6≡ 0 in Ω. If ϕ |∂Ω≡ c1 andc1 = 0, ϕ ≡ 0 by to Lemma 4.3 and consequently ϕ ≡ 0. So, we may assumec1 < 0. Hence, the nodal line ϕ = 0 is away from ∂Ω and divides Ω into twocomponents Ω+ = x | ϕ(x) > 0 and Ω− = x | ϕ(x) < 0. Furthermore,by Lemma 4.3, Ω+ is simply-connected and satisfies

Ω+ev(x)dx ≥ 4π ≥

Ω−

ev(x)dx.(4.29)

Set Ω+ ⊂ Ω+ = x | ϕ(x) > c1 and Ω− = x | ϕ(x) < c1. We discuss twocases separately.Cases 1. Assume Ω− is an empty set.

In this case we can do the symmetrization for ϕ in Ω as in the proof ofLemma 4.3, that is, for any t > c1, we set Br(t) to be the open ball of center0 and radius r(t) such that

Br(t)ev(x)dx =

ϕ>teU(x)dx,

where U(x) is defined as in Lemma 4.3. As before, r(t) is continuous, strictlydecreasing in t and limt↓c1 R(t) = +∞, because

Ω ev(x)dx = 8π. The sym-

metrization ϕ∗(x) = ϕ∗(|x|) is defined as the same as before. By Lemma 4.2and the coarea formulas, we obtain as before

R2| 5 ϕ∗|2dx ≤

Ω| 5 ϕ|2dx =

Ωevϕ2(x)dx =

R2eU ϕ∗2

dx,(4.30)

and ∫

R2eU ϕ∗(x)dx =

Ωev(x)ϕ(x)dx = 0.(4.31)

Set

K∗ = inf

R2| 5 ψ|2dx | ψ(x) is radially symmetric,(4.32)

R2eU(x)ψ(x)dx = 0 and

R2eU(x)ψ2(x)dx = 1

.

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Note that

R2eU(x)ψ(x)dx

≤(∫

R2eU(x)ψ2(x)dx

)12(∫

R2eU(x)dx

)12

.

Hence, condition (4.31) also holds for any minimizer of (4.32). Set ψ∗ to bethe minimizer of (4.32) and ψ∗ satisfies

∆ψ∗ +K∗eU(x)ψ∗ = 0 in R2,∫

R2 eU(x)ψ∗dx = 0.(4.33)

By (4.30), K∗ ≤ 1. Therefore, ψ∗ changes the sign once and only once,otherwise it yields a contradiction to Lemma 4.3. Let ξ0 be the zero of ψ∗

and we may assume ψ∗(r) > 0 if r < ξ0 and ψ∗(r) < 0 if r > ξ0. By (4.31),

rψ∗′(r) =∫ r

0eU(s)ψ∗(s)sds(4.34)

=∫ ∞

reU(s)ψ∗(s)sds < 0

for r ∈ (ξ0,∞). Therefore, ψ∗′(r) is decreasing for r ≥ ξ0 and rψ∗′(r) → 0 asr → +∞. Clearly, (4.33) yields

|rψ∗′(r)| ≤(

∫ ∞

reU(s)ψ∗2

(s)sds

)12(

∫ ∞

reU(s)sds

)12

≤ c r−1

for large r. Hence limr→+∞ ψ∗(r) exists and

limr→+∞

ψ∗(r) < 0(4.35)

Let ψ(r) = 8−|x|2

8+|x|2. Then ψ satisfies

∆ψ + eUψ(r) = 0.(4.36)

Together with (4.33), we have

r

(

ψ∗

ψ(r)

)′

=(1 −K∗)

ψ2(r)

∫ r

0eU(s)ψ∗(s)ψ(s)sds.

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If ξ0 <√

8 where ψ(√

8) = 0, then ψ∗(r)ψ(r)

is increasing for r ∈ [0, ξ0]. Clearly,it yields

0 <ψ∗(0)

ψ(0)<ψ∗(ξ0)

ψ(ξ0)= 0,

a contradiction. Thus, ξ0 ≥√

8. Similarly, we have

limR→+∞

R

(

ψ∗

ψ

)′

(R)ψ2(R) − r

(

ψ∗

ψ

)′

ψ2(r) = (1 −K∗)∫ ∞

reU(s)ψ∗(s)ψ(s)sds.

Sincelim

R→+∞R[ψ∗′(R)ψ(R) − ψ′(R)ψ∗(R)] = 0,

We have

−(

ψ∗

ψ

)′

ψ2(r) = (1 −K∗)∫ ∞

reU(s)ψ∗(s)ψ(s)sds.

If ξ0 >√

8, then ψ∗

ψ(r) is decreasing for r ≥ ξ0. Then it yields

0 =ψ∗(ξ0)

ψ′(ξ0)> lim

r→+∞

ψ∗(r)

ψ(r)= − lim

r→+∞ψ∗(r),

a contradiction to (4.35). Therefore, we conclude ξ0 =√

8 and ψ∗(r)ψ(r) > 0for all r 6=

√8. Again,

0 = limr→+∞

(ψ∗(r)′ψ(r) − ψ′(r)ψ∗(r))r

= (1 −K∗)∫ ∞

0eU(s)ψ∗(s)ψ(s)sds

yields K∗ = 1, and (4.30) becomes an equality. Therefore, all the inequalitiesof (4.10) turn out to be equalities. In particular,

l2ϕ = t =1

2

(

8π −∫

Ωt

ev(x)dx

)

Ωt

ev(x)dx(4.37)

holds for almost everywhere t. Since ϕ is continuous on Ω, we have

l2ϕ = t ≥ c > 0

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for some constant c > 0 and for t is close to c1. On the other hand, for anyε > 0,

8π −∫

Ωt

ev(x)dx ≤ ε

holds if t is sufficiently close to c1. Thus, it yields a contradiction to (4.37)when t is close to c1. This contradiction proves that Ω− is non-empty.Cases 2. Ω− is not empty. In this case, we still do the symmetrization forΩ+ as before. Set R0 be

ϕ>c1ev(x)dx =

BR0

eU(x)dx(4.38)

But for t < c1, we symmetrize ϕ < t as

R2\Br(t)

eU(x)dx =∫

ϕ(x)<tev(x)dx(4.39)

Since∫

Ω evdx = 8π, we have limt↑c1 r(t) = R0. Set

ϕ∗∗(r) = inft | x ∈ R2\Br(t).

for r > R0 andlimr↓R0

ϕ∗∗(r) = lims↑R0

ϕ∗(s)(4.40)

By the isoperimetric inequality (4.2), the symmetrization yields

BR0

| 5 ϕ∗(x)|2dx ≤∫

Ω+| 5 ϕ|2dx(4.41)

as (4.30). For t < c1, we note that ϕ = t bounds the domain x | ϕ < twhich obiviously is not a simply-connected domain. Thus, by (4.2)

(

ϕ=te

v2

)2

>1

2

(

8π −∫

ϕ<tevdx

)(

ϕ<tevdx

)

for t < c1. Then (4.10) yields for t < c1

− d

dt

Ωt

| 5 ϕ|2dx =∫

ϕ=t| 5 ϕ|ds

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>1

2

(

8π −∫

ϕ<tevdx

)(

ϕ<tevdx

)[

− d

dt

Ωt

ev(x)dx

]−1

=1

2

(

8π −∫

R2\Br(t)

eUdx

)(

R2\Br(t)

eUdx

)[

− d

dt

R2\Br(t)

eU(x)dx

]−1

= − d

dt

R2\Br(t)

| 5 ϕ∗∗|2dx.

Integrating from c1 to infΩ ϕ, we obtain

Ω| 5 ϕ|2dx−

Ω+| 5 ϕ|2dx ≥

R2\BR0

| 5 ϕ∗∗|2dx.(4.42)

By (4.40), ϕ(r) is continuous at R0, where ϕ is defined by

ϕ(r) =

ϕ∗ if r < R0,

ϕ∗∗ if r > R0.

Together with (4.41) and (4.42), it yields

R2| 5 ϕ|2dx <

Ω| 5 ϕ|2dx =

Ωevϕ2dx =

R2eU(ϕ)2dx.(4.43)

Clearly, ϕ also satisfies

R2eU ϕ(x)dx =

Ωevϕ(x)dx = 0.

As before, we set

K∗ =

R2| 5 ψ|2dx

ψ(x) is radially symmetric,

R2eUψ(x)dx = 0 and

R2eU(x)ψ2(x) = 1

.

The minimizing is exactly the same one as (4.32). By the previous argument,we have K∗ = 1. On the other hand, K∗ < 1 by (4.43). This contradictionshows that there are no null eigenfunctions of the linearized equation exceptthe trivial one.

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Since the linearized equation has only the trivial solution, it is not hardto prove the estimate for the inverse operator. For example, we can showthat

‖ϕ‖H10 (Ω) + ‖ϕ‖L∞(Ω) ≤ c ‖L(ϕ)‖L∞(Ω),(4.44)

where L denotes the linearized operator of (4.25). Note that both ‖ϕ‖H10 (Ω)

and ‖ϕ‖L∞(Ω) is invariant under the conformal transformation. By using theimplicit function theorem and (4.44), we can prove that for any solution u

of (1.1) with ρ = 8π, there is a solution uρ(x) of (1.1) with small |ρ − 8π|and limρ→8π uρ(x) = u(x). Thus, the uniqueness of (1.1) with ρ = 8π followsreadily from Theorem 4.1. Q.E.D.

5 Domains with D(p) = 0

In this section, we discuss critical points of γ with D(p) ≤ 0. Throughoutthis section, Ω is always assumed to be simply-connected. Now we are in theposition to prove Theorem 1.7.

Proof of Theorem 1.7. We first note that if the nondegenerate conditionis not satisfied, then

D(p) =∞∑

n≥3

n2|an|2n− 2

− |a1|2 > 0

unless an = 0 for n ≥ 4. But, if an = 0 for n ≥ 4, f(z) = z + 13z3 after

a rotation and a scaling. Clearly, f ′(±i) = 0, and it implies that Ω has acusp at its boundary point ± 2

3i and its inner angle at ± 2

3i are 2π. Call Ω0

to be this particular domain. Note that Ω0 is not a piecewisely C2 domaindiscussed in this paper. Hence we conclude that if D(p) ≤ 0 then p is anondegenerate critical point.

Suppose D(p) < 0. Since p is a nondegenerate critical point, there existsa sequence of solutions uk of (1.1) with ρ = ρk such that uk blows up atp. By Theorem 1.4, we know ρk < 8π. Thus, by the uniqueness theoremof Suzuki, uk is the minimizer of the nonlinear functional Jρk

. Clearly, theblowup point p must be a maximum point of γ. Now suppose that thereexists another maximum point q 6= p. If D(q) < 0, then there exists another

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sequence of solutions vk with ρk < 8π, which blows up at q. For k large, wehave vk 6= uk. Then we obtain a contradiction to the uniqueness theorem ofSuzuki. If D(q) = 0, we can perturb Ω slightly to Ω′ such that q is also acritical point of γΩ′ and DΩ′(q) < 0. We can perturb the domain Ω by usingthe conformal map. Set f(w) be the conformal mapping from the unit ball toΩ with f(0) = q. Denote fη(w) = q +

∑∞n≥3 anη

nwn + (a1η)w, for 0 < η < 1.Let Ωη be the image of the unit B1 under the conformal fη. Clearly, Ωη is asmooth domain Ωη → Ω as η → 1. By (1.16), DΩη(q) < D(q) = 0. So, wechoose Ω′ to be one of Ωη for small 1− η > 0. Note that because a2 = 0, q isalso a critical point of Ω′. Since Ω′ is a small perturbation of Ω, there existsanother critical point p′ of γΩ′ near by p such that DΩ′(p′) < 0 (because p isa nondegenerate critical point of γΩ). Since p′ 6= q, it violates the conclusionof the previous step. The last case is D(q) > 0. Then by the computation in

the next section, we have I8π(Ω) > 1 + supΩ γ(x) + log |B1||Ω|

. Then I8π(Ω) canbe attained by some extremal function v. by Theorem 1.6 and the implicitfunction theorem, there exists a sequence of solutions v(x; ρ) of (1.1) withv(x; ρ) → v(x) as ρ ↑ 8π. Thus, v(x; ρk) 6= uk(x) because uk blows up at p.This violates the uniqueness theorem for ρ < 8π. This finishes the proof ofthe uniqueness of the maximum point p with D(p) < 0.

Now suppose D(p) = 0. We claim that there exists a blowup sequenceof solutions uk of (1.1) with ρk ↑ 8π such that p is the blowup point. Theclaim is proved by using Theorem 1.6, the uniqueness theorem for ρ = 8π.To see it , we can perturb the domain Ω to obtain a sequence of domainsΩj → Ω such that for each j, p ∈ Ωj and p is a critical point of γj = γΩj

withDj(p) = DΩj

(p) < 0. By the previous step, we know p is the unique maximumpoint of γj and for each j and any ρ ∈ [0, 8π), there exists an unique solutionuj(x; ρ) of (1.1). Since uj(x; ρ) is unique, uj(x; ρ) is smoothly depending onthe parameter ρ for each j. Since Dj(p) < 0, supΩj

uj(x; ρ) → +∞ as ρ ↑ 8πfor each j and the blowup point uj(x, ρ) is p only, as ρ ↑ 8π. Thus, for anylarge constant C > 0 and for any j, there exists a ρj ∈ [0, 8π) such that

supΩj

uj(x; ρj) = C, and(5.1)

supB δ0

2

(p)uj(x; ρj) = C > 2 sup

Ωj\Bδ0(p)

uj(x; ρj),(5.2)

where δ0 is a small positive number such that 4δ0 < dist (p, q) for any criticalpoint q 6= p of γ. By passing to a subsequence of j, uj(x; ρj) uniformly

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converges to a solution u(x, C) of (1.1) with ρ(C) = limj→+∞ ρj ≤ 8π. Bythe uniqueness theorem of (1.1) with ρ = 8π, equation (1.1) for Ω possessesone solution at most. Hence if C is chosen to be large, then ρ(C) < 8π. Byletting C = k → +∞, we then construct a sequence of blowup solutions ukof (1.1) with ρ = ρk ↑ 8π which blows up at p. By (5.2), the blowup pointof uk must be p. Hence the claim is proved. Since p is a maximum point ofγj(x), by passing to the limit, p is a maximum point of γ(x).

We remain to prove p is the unique maximum point. suppose q 6= p isanother maximum point. Then by the previous step, D(q) ≥ 0. The caseD(q) > 0 is also excluded by the same argument before. So, we assumeD(q) = 0. Then by the claim, we have two different sequence of blowup solu-tions of (1.1) with ρ ↑ 8π. This violates Theorem 1.6 again. Hence Theorem1.7 is proved. Q.E.D.

6 Extremal functions

Proposition 6.1. Suppose that Ω is bounded piecewisely C2 simply con-nected domain and u(x; ρ) is the minimizer of Jρ with ρ ∈ [0, 8π). Then thefollowings are equivalent.

(i) u(x; ρ) is uniformly bounded for x ∈ Ω and ρ ∈ [0, 8π),

(ii) Ω is of type C, and

(iii) Equation (1.1) with ρ = 8π possesses a solution.

Proof. Obviously, (i) ⇒ (ii) and (ii) ⇒ (iii). Now suppose (1.1) possessesa solution u for ρ = 8π. Then by Theorem 1.6 and the implicit functiontheorem, for |ρ− 8π| small, equation (1.1) possesses a solution u(x; ρ) whichtends to u as ρ ↑ 8π By the uniqueness theorem, u(x; ρ) is the minimizer ofJρ for ρ < 8π. Hence (iii) ⇒ (i) is proved.

Lemma 6.2. Let Ω be a bounded domain of R2 and p be a critical point ofγ. If D(p) > 0, then I8π(Ω) > 1 + maxΩ γ(x) + log( |B1|

|Ω|).

Proof. Let f be the conformal map from the unit ball B1 onto Ω, withf(0) = p. For any ε > 0, we set

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vε(z) = 2 log

(

1 + ε

ε+ |z|2)

(6.1)

for |z| < 1, and denote uε as a test function by

uε(y) = vε(f−1(y))

for y ∈ Ω. Then

1

16π

Ω| 5 uε(y)|2dy =

1

16π

B1

| 5 vε|2dz(6.2)

= 2∫ 1

0

r3

(ε+ r2)2dr

= log

(

1 + ε

ε

)

− 1

1 + ε

and,

Ωeuε(y)dy =

B1

|f ′(z)|2 (1 + ε)2

(ε+ |z|2)2dz

= 2π

(

|a1|2∫ 1

0

(1 + ε)2r

(ε+ r2)2dr +

∞∑

n=3

|an|2n2(1 + ε2)∫ 1

0

r2n−1

(ε+ r2)2dr

)

= π

(

|a1|2(1 + ε)

ε+

∞∑

n=3

n2

n− 2|an|2 +O(ε)

)

.

Thus,

J8π(uε) = log|B1||Ω| + log

(

1

π

B1

|f ′(z)|2evε(z)dx

)

− log(ε+ 1)2

ε

= log|B1||Ω| + log

[

|a1|2ε

+

(

|a1|2 +∞∑

n=3

n2

n− 2|an|2

)

+O(|ε|)]

+ logε

(ε+ 1)+

1

1 + ε

= log|B1||Ω| + log

[

|a1|2 +

(

∞∑

n=3

n2

n− 2|an|2

)

ε+O(|ε|2)]

+1 − ε+O(ε2)

= log|B1||Ω| + 1 + log |a1|2 + |a1|−2D(p)ε+O(|ε|2).

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Since γ(p) = 12π

log |a1|, we have

J8π(uε) = 4πγ(p) + 1 + log|B1||Ω| + |a1|−2D(p)ε+O(|ε|2)

> 1 + 4πγ(p) + log|B1||Ω| ,

provided that ε is sufficiently small and D(p) > 0. Q.E.D.

Now we can give proofs of Theorem 1.5 and Theorem 1.6.

Proof of Theorem 1.5. By lemma 6.2, if there is a maximum point p ofγ(x) with D(p) > 0, then I8π(Ω) > 1 + 4πmaxΩ γ(x) + log |B1|

|Ω|. Then by the

”if” part of Theorem 1.1 (which was already proved in [5,Theorem 7.1]), thesupreman of J8π can be attained.

Now suppose all the maximum points p of γ(p) satisfy D(p) ≤ 0. Thenby Theorem 1.7, there is only one unique maximum point. By Theorem 1.8,there exists a sequence of solutions u(x; ρ) of (1.1) with ρ < 8π such thatu(x; ρ) blows up at p exactly. Then by (i) of Proposition 6.1, the supremanof J8π can not be attained. Q.E.D.

Proof of Theorem 1.6. Now suppose that the supreman of J8π can beattained. Then by Theorem 1.5, D(p) > 0 for all the maximum points p of

γ. Applying Lemma 6.2, we then have I8π(Ω) > 1 + 4πmaxΩ γ + log |B1||Ω|

.Q.E.D.

7 Final discussions

In this section, we will give several examples to interprete our theorems.First, we would like to apply Theorem 1.5 to the domain Ω of regular n-polygon.

Proposition 7.1. Let Ω be the regular n-polygon. Then Ω is not of typeC.

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Proof. Let the origin be the center of Ω. Since Ω is convex, 0 is the uniquecritical point of γ. By the Schwartz-Christoffel formulas, we have

f ′(z) = (1 − zn)−2n =

∞∑

k=0

2n( 2n

+ 1) . . . ( 2n

+ k − 1)

k!znk.

See [1]. Then

Dn(0) =∞∑

k=1

(

2n( 2n

+ 1) . . . ( 2n

+ k − 1)

k!

)2(1

nk − 1

)

−1.

Clearly, Dn(0) ≤ D3(0). Let Ak =

(

23( 23+1)...( 2

3+k−1)

k!

)2

13k−1

. Since Ak+1

Ak=

1− 15k+119(k+1)2

≤ (k+1k+2

)53 , we have Ak ≤ C(k+1)−

53 for k = 2, . . ., where C = 5

273

23 .

Therefore,

D3(0) ≤ 2

9+

5

273

23

∞∑

k=2

(k + 1)−53 − 1

=5

18(3

2)

23 − 7

9< 0,

where∑∞k=3 k

− 53 ≤ 3(1

2)

53 is used. By Theorem 1.5, Ω is not of type C.Q.E.D.

Now suppose Ω to be a rectangle with sides 0 < a ≤ b. Proposition 7.1tells us that if a = b, then Ω is not of type C. On the other hand, if b

ais

sufficiently small, then Ω is of type C. In general, we have the following result.

Proposition 7.2. There exists a constant 0 < η0 < 1 such that Ω is of typeC if and only if b

a< η0.

Proof. Letf ′θ(z) = (1 − z2)−

12 (1 − e−2iθz2)−

12

for |z| < 1, where θ is a constant with 0 < θ < π2. By the Christoffel-

Schwartz formulas, fθ maps the unit ball onto a rectangle Ω(θ). Let a(θ) andb(θ) denotes the two sides of Ω(θ). We claim

a(θ)

b(θ)is an increasing function of θ.(7.1)

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Note that

a(θ) =∫ θ

0

|1 − e2iϕ| 12 |1 − e2i(ϕ−θ)| 12=

1

4

∫ θ

0

dϕ√

sinϕ sin(θ − ϕ)

=1

4

∫ θ2

−θ2

dϕ√

sin(ϕ+ θ2) sin( θ

2− ϕ)

=1

4

∫ θ2

−θ2

dϕ√

sin2 θ2− sin2 ϕ

.

By changing the variable τ = sinϕ, we have

a(θ) =1

4

∫ τ0

−τ0

dτ√

(1 − τ 2)(τ 20 − τ 2)

=1

4

∫ 1

−1

dτ√

(1 − τ 20 τ

2)(1 − τ 2),

where τ0 = sin θ2. Obviously, a(θ) is increasing in θ.

The area of the rectangle Ω(θ) is

A(θ) =∫

B1

dxdy

|1 − z2||1 − e−2iθz2| =∫

B1

dxdy

|1 − z2||e2iθ − z2| .

We want to prove A(θ) is decreasing in θ. Set 0 < θ1 < θ2 ≤ π2, the line

L = z = tei(θ1+θ2) | |t| ≤ 1 decomposes the unit ball into two regions Ω1

and Ω2, where e2iθ1 ∈ Ω1 and e2iθ2 ∈ Ω2. Then

A(θ1) − A(θ2)(7.2)

=∫

B1

(|1 − z2||e2iθ1 − z2|)−1 − (|1 − z2||e2iθ2 − z2|)−1dxdy

=1

2

B1

1

|z|

[(1 − z)|e2iθ1 − z|]−1 − [|1 − z||e2iθ2 − z|]−1

dxdy

=1

2

Ω1

|z|−1(|1 − z|)−1

(|e2iθ1 − z|)−1 − (|e2iθ2 − z|)−1

dxdy

−1

2

Ω2

|z|−1(|1 − z|)−1

|e2iθ2 − z|)−1 − (|e2iθ1 − z|)−1

dxdy.

Let z∗ be the reflection point of z with respect to the line L. Then

|e2iθ2 − z| = |e2iθ1 − z∗|, and,

|e2iθ1 − z| = |e2iθ2 − z∗|

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for z ∈ Ω2. Therefore

A(θ1) − A(θ2)

=1

2

Ω1

1

|z|

1

|1 − z| −1

|1 − z∗|

(|e2iθ1 − z|)−1 − (|e2iθ2 − z|)−1

dxdy > 0,

where 1 ∈ Ω1 and |1 − z| < 1 − |z∗| for z ∈ Ω1.

Write a(θ)b(θ)

= a2(θ)A(θ)

. Then (7.1) is proved. Obviously, limθ→π2

a(θ)b(θ)

= 1 and

limθ→0a(θ)b(θ)

= 0.Set

D(θ) =∫

B1

(

|f ′θ(z)|2|z|4 − 1

|z|4)

dxdy −∫

Bc1

1

|z|4dxdy

For 0 < θ1 < θ2, we have

D(θ1)−D(θ2) =∫

B1

1

|z|4

(|1− z2||e2iθ1 − z2|)−1− (|1− z2||e2iθ2 − z2|)−1dxdy.

By using the same argument of (7.2), we can prove D(θ1) > D(θ2). HenceD(θ) is decreasing in θ. For θ = π

2, Ω(π

2) is a cube. Hence D(π

2) < 0. On

the other hand, it is known that D(θ) > 0 if θ is small. Therefore, there

exists θ0 ∈ (0, π2) such that D(θ) > 0 if and only if θ < θ0. Let η0 = a(θ0)

b(θ0).

Proposition 7.2 follows readily from Theorem 1.5. Q.E.D.

Next, we consider the domain Ωh to be a dumbell which consists of twodisjoint balls B1 and B2 connected with a tube of small width h > 0. Letr1 ≤ r2 be the radius of B1 and B2 respectively.

Proposition 7.3. Let Ωh be the domain described above.

(i) If r1 < r2, then for small h > 0, Ωh is not of type C, and

(ii) If r1 = r2 and Ωh is assumed to be symmetric with respect to x2-axis,then Ωh is of type C.

Proof. Let p1 and p2 are the center of B1 and B2 respectively. Gh andγh denote the Green function and the regular part of Gh. Obviously, γh(x)converges to the regular part of B1 (and of B2) uniformly in any compact setx of B1 (and of B2). Thus, there exists a local maximum of γh at p1,h and

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p2,h with limh→0 pi,h = pi for i = 1, 2. We use the formulas (1.12), instead ofthe series sum (1.16), to compute D(pi,h) for i = 1, 2. For example at p1,h,the associated function Hh converges to

H(y, 0) =

0 if y ∈ B1

r−41 |y − p1|4 − 1 if y ∈ B2

as h ↓ 0. Thus,

B1∪B2

H(y, 0)

|y − p1|4dy −

(B1∪B2)c

dy

|y − p1|4= πr−4

1 r22 −

Bc1

dy

|y − p1|4= π(r2

2 − r21)r

−41 > 0.

Hence D(p1,h) > 0 for Ωh. Similarly, we can compute

limh↓0

D(p2,h) = π(r21 − r2

2)r−42 < 0.

By Theorem 1.5, p2,h must be the maximum point and Ωh is not of type Cprovided that h is sufficiently small. This proves (i).

For r2 = r1, the shape of the connection tube might affect the type ofdomains. We consider only the symmetric one. Obviously, p2,h and p1,h arethe maximum points of γh for small h where p1,h and p2,h tends to the centersof B1 and B2 respectively as h ↓ 0. Since those maximum point p of γ withD(p) ≤ 0 is unique, we see D(p1,h) = D(p2,h) must be positive. By Theorem1.5, Ωh is of type C. This proves (ii). Q.E.D.

In this paper, we have already seen that Theorem 1.6 plays an impor-tant role. In general, we do not expect it holds for non-simply connecteddomains. Nevertheless, it still holds for an annulus domain. Note for thedomain Ωa = x | a < |x| < 1. We have proved that I8π(Ωa) is attained bysome extremal functions. See [10].

Proposition 7.4. Let Ωa = x | a < |x| < 1. Then any solution u of(1.1) is radially symmetric for any β ∈ (0, 8π]. Moreover, the solution isunique.

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Proof. Set ϕ(x) = ∂u∂θ

. Then ϕ satisfies

∆ϕ+ ρeuϕ∫

Ωaeudx

= 0 in Ωa,

ϕ = 0 on ∂Ωa.

If ρ ≤ 8π, we will apply the isoperimatic inequality to show that ϕ ≡ 0.Assume ϕ 6≡ 0. Our proof is based on the simple observation: for anyr ∈ (a, 1), there exists at least two zeros of ϕ on the cricle x | |x| = r. Thusϕ = 0 must intersect with each connected component of the boundary ∂Ωa

and ϕ has at least two critical points. Now suppose that ϕ has a critical pointat some point p ∈ ϕ = 0, then by the simple geometry of the plane, thenodal line ϕ = 0 must enclose two simply connected domains. Then we canfollows the symmetrization argument of Lemma 4.3 to obtain a contradictionif ρ ≤ 8π. This proves ϕ ≡ 0 and then u must be radially symmetric. Theuniqueness of radial solution (1.1) for any β ∈ R has been proved in [5].Q.E.D.

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Sun-Yung A. Chang Chiun-Chuan ChenDepartment of Mathematic Department of MathematicsPrinceton University National Taiwan [email protected] Taipei, Taiwan

[email protected] LinDepartment of MathematicsNational Chung-Cheng UniversityMinghsiung, [email protected]

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